I wanted to confirm my understanding of the change of basis matrix and its notation and would appreciate if someone could follow my thinking and verify this please. Given two basis, $A$ and $B$ in the vector space $V \in \Bbb{R}^n$ and the linear maps $\phi_A$ and $\phi_B$ which take the standard basis:
$\phi_A(e_i) = \alpha_i $ and $\phi_B(e_i) = \beta_i $
Where $\alpha_i$ and $\beta_i$ are the basis vectors of the basis $A$ and $B$ for $i = 1,...,n$
For a vector $x$ in $V$ whose coordinate vector wrt $A$ and $B$ is $x_A$ and $x_B$ respectively we have
$[L]_E^Ax_E = x_A$ and $[L]_E^Bx_E = x_B$
where $[L]_E^A$ and $[L]_E^B$ are the matrices, associated with $\phi_A$ and $\phi_b$ that perform matrix multiplication.
So by assuming that the linear maps are non-singular, taking the inverse and equating $x_E$ we should get:
$x_A = [L]_E^A [L]_B^Ex_B $
which is the coordinate representation of the vector $x$ wrt the basis $A$ and $[L]_E^A [L]_B^E$ is the transformation matrix, $[L]_B^A$
Ultimately we have $ x_A = [L]_B^Ax_B $
Is my understanding and formalisation here correct or have I gone wrong somewhere?
Yes your reasoning is fine, for the notation in matrix form we have
$$x_A=[L]_E^A \,x_E \quad x_B=[L]_E^B \,x_E$$
therefore
$$x_A=[L]_E^A \,x_E=L]_E^A ([L]_E^B)^{-1}\,x_B=L]_E^A [L]_B^E\,x_B=[L]_B^A \,x_B$$
with $[L]_B^A=L]_E^A [L]_B^E$.